Question: Find $\lim_{x\to -3}\dfrac{4x^2+12x}{x^2+4x+3}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $-12$ (Choice C) C $6$ (Choice D) D The limit doesn't exist
Solution: Substituting $x=-3$ into $\dfrac{4x^2+12x}{x^2+4x+3}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{4x^2+12x}{x^2+4x+3}$ can be simplified as $\dfrac{4x}{x+1}$, for $x\neq -3$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{4x^2+12x}{x^2+4x+3}=\dfrac{4x}{x+1}$ for all $x$ -values in the interval $(-4,-2)$ except for $x=-3$. Therefore, $\lim_{x\to -3}\dfrac{4x^2+12x}{x^2+4x+3}=\lim_{x\to -3}\dfrac{4x}{x+1}=6$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -3}\dfrac{4x^2+12x}{x^2+4x+3}=6$.